Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Segment {
  double l, r;
};
vector<Segment> segments;
vector<long long> segSum;
double findOverlap(double l, double r) {
  int s1l = 0, s1r = segments.size() - 1, s1m;
  while (s1l < s1r) {
    s1m = (s1l + s1r) / 2;
    if (segments[s1m].r < l)
      s1l = s1m + 1;
    else
      s1r = s1m;
  }
  int s2l = 0, s2r = segments.size() - 1, s2m;
  while (s2l < s2r) {
    s2m = (s2l + s2r) / 2;
    if (segments[s2m].r < r)
      s2l = s2m + 1;
    else
      s2r = s2m;
  }
  double s = s2l > 0 && s2l > s1l ? segSum[s2l - 1] - segSum[s1l] : 0;
  s += max(0.0, min(r, segments[s1l].r) - max(l, segments[s1l].l));
  if (s1l != s2l)
    s += max(0.0, min(r, segments[s2l].r) - max(l, segments[s2l].l));
  return s;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  double sy, a, b;
  int q, n;
  cin >> sy >> a >> b >> n;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    segments.push_back({.l = (double)l, .r = (double)r});
    segSum.push_back((i > 0 ? segSum[i - 1] : 0) + r - l);
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x, y;
    double l, r;
    cin >> x >> y;
    l = a - sy / (y - sy) * (x - a);
    r = b - sy / (y - sy) * (x - b);
    double z = findOverlap(l, r) / (r - l) * (b - a);
    printf("%.15f\n", z);
  }
  return 0;
}